Problem: Let $f(x) = 6x^{2}+10x-8$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Solution: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $6x^{2}+10x-8 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 6, b = 10, c = -8$ $ x = \dfrac{-10 \pm \sqrt{10^{2} - 4 \cdot 6 \cdot -8}}{2 \cdot 6}$ $ x = \dfrac{-10 \pm \sqrt{292}}{12}$ $ x = \dfrac{-10 \pm 2\sqrt{73}}{12}$ $x =\dfrac{-5 \pm \sqrt{73}}{6}$